Subject: Re: Difficult optimization in MOPAC

From: [email-masked] (Dr. James Stewart)
Date: Thu, 27 Jan 94 08:48:22 PST

I thank everyone who sent me suggestions on how to optimize the geometry
of C6H5-COO radical using RHF techniques.   The best suggestion was given
by Dr John McKelvey.  His solution to the problem is given in the following
ARC file from MOPAC 93.  The GNORM was reduced to below the pre-specified
limit of 0.01, which is definitive proof of the validity of the optimization.

About 5-7 days had been spent on attempting the optimization, without any success.
Within a day of posting the problem, I had received the correct answer.  This
demonstrates the usefulness of the Net, and the skill of its users.

Jimmy Stewart

                     SUMMARY OF   PM3   CALCULATION

                                                            MOPAC  93.00

  C7  H5  O2 
                                                       Thu Jan 27 09:28:47 1994
    GNORM=0.01 GRADIENTS  PRECISE  PM3 SHIFT=50 EF C.I.=(3,2)
 C6H5-COO(.)

     GEOMETRY OPTIMISED USING EIGENVECTOR FOLLOWING (EF).     
     SCF FIELD WAS ACHIEVED                                   

          HEAT OF FORMATION       =       -10.726466 KCAL =    -44.87954 KJ
          ELECTRONIC ENERGY       =     -6178.829220 EV      STATE: DOUBLET B2  
          CORE-CORE REPULSION     =      4687.367953 EV
          GRADIENT NORM           =         0.009496
          DIPOLE                  =         4.39069 DEBYE    SYMMETRY:       C2v 
          NO. OF FILLED LEVELS    =        22
          AND NO. OF OPEN LEVELS  =         1
          CONFIGURATION INTERACTION WAS USED
          IONIZATION POTENTIAL    =        10.354278 EV
          HOMO (SOMO) LUMO (EV)   =       -10.439 ( -7.018) -0.921
          MOLECULAR WEIGHT        =       121.115
          SCF CALCULATIONS        =        17
          COMPUTATION TIME = 12 MINUTES AND  10.633 SECONDS

          FINAL GEOMETRY OBTAINED                                    CHARGE
    GNORM=0.01 GRADIENTS  PRECISE  PM3 SHIFT=50 EF C.I.=(3,2)
 C6H5-COO(.)
 
  C    0.00000000  0    0.0000000  0    0.0000000  0    0    0    0     -0.0563
  C    2.78147112  1    0.0000000  0    0.0000000  0    1    0    0     -0.1536
  C    1.39164098  1   60.0624509  1    0.0000000  0    1    2    0     -0.1172
  C    1.39164180  1   60.0622252  1  179.9992100  1    1    2    3     -0.1172
  C    1.39738087  1   60.0816056  1   -0.0001412  1    2    1    3     -0.0343
  C    1.39737820  1   60.0816087  1 -179.9989763  1    2    1    3     -0.0343
  H    1.09548486  1  179.9966649  1 -103.5372856  1    1    2    3      0.1071
  H    1.09537441  1  120.0286616  1 -179.9985171  1    3    1    2      0.1124
  H    1.09537460  1  120.0285221  1  179.9985854  1    4    1    2      0.1124
  H    1.09678871  1  120.1471184  1  179.9997238  1    5    2    1      0.1195
  H    1.09678839  1  120.1472596  1 -179.9983554  1    6    2    1      0.1195
  C    1.47024274  1  179.9997536  1    3.4455962  1    2    1    3      0.4514
  O    1.26492843  1  130.2209371  1 -179.9902413  1   12    2    5     -0.2548
  O    1.26492930  1  130.2194911  1    0.0089828  1   12    2    5     -0.2548